Bash arithmetic expression's side effects not executed

Question

$ declare -i i=0
$ for j in {0..2}; do echo "${j} $((i++))"; done
0 0
1 1
2 2
$ for j in {0..2}; do echo "$(echo "${j} $((i++))")"; done
0 3
1 3
2 3
$

Why i doesn't get incremented in the 2nd for loop?

(Yes, I know that there's a workaround.)

Answer 1

It gets incremented in the subshell created by the $(command substitution). When that process exits, the modified value is lost.

Here are similar ways of seeing the same effect:

i=0

bash -c 'let i++'    # Subprocess
( let i++  )         # Explicit subshell
let i++ & wait       # Backgrounded process
: <( let i++ )       # Process substitution
let i++ | cat        # Pipeline

echo "$i"            #  Still 0